find-0-1-dx-1-x-n-interms-of-digamma-

Question Number 174117 by Mathspace last updated on 25/Jul/22

f i n d \int_{0}^{1} \frac{d x}{1 + x^{n}} i n t e r m s o f

ψ (d i g a m m a)

Answered by Mathspace last updated on 25/Jul/22

\int_{0}^{1} \frac{d x}{1 + x^{n}} = \int_{0}^{1} \sum_{p = 0}^{\infty} {(- 1)}^{p} x^{n p} d x

= \sum_{p = 0}^{\infty} {(- 1)}^{p} \int_{0}^{1} x^{n p} d x

= \sum_{p = 0}^{\infty} \frac{{(- 1)}^{p}}{n p + 1}

= \sum_{k = 0}^{\infty} \frac{1}{2 n k + 1} - \sum_{k = 0}^{\infty} \frac{1}{(2 k + 1) n + 1}

= \sum_{k = 0}^{\infty} (\frac{1}{2 n k + 1} - \frac{1}{2 n k + n + 1})

= n \sum_{k = 0}^{\infty} \frac{1}{(2 n k + 1) (2 n k + n + 1)}

= \frac{n}{4 n^{2}} \sum_{k = 0}^{\infty} \frac{1}{(k + \frac{1}{2 n}) (k + \frac{n + 1}{2 n})}

= \frac{1}{4 n} (ψ (\frac{n + 1}{2 n}) - ψ (\frac{1}{2 n})) \times \frac{1}{\frac{n + 1}{2 n} - \frac{1}{2 n}}

= \frac{1}{2 n} {ψ (\frac{1}{2 n} + \frac{1}{2}) - ψ (\frac{1}{2 n})}

Commented by aleks041103 last updated on 25/Jul/22

Y e s, t h i s i s c o r r e c t! I m a d e a n e x t r e m e l y

b i g m i s t a k e o f d o i n g w h a t e v e r I w a n t w i t h

n o n c o n v e r g i n g i n t e g r a l s \dots

Commented by Mathspace last updated on 25/Jul/22

y o u r m e t h o d i s c o r r e c t b y y o u h s v e c o m m i t e d

a e r r o r o f c a l c u l u s n e v e r m i n d s i r